# Braket operation

1. Jul 26, 2012

### intervoxel

Is this operation legal?

$\langle A_1|BA|A_2\rangle=A_2\langle A_1|B|A_2\rangle$

If so, please elaborate on that.

Thanks.

2. Jul 26, 2012

### vanhees71

If $|A_2 \rangle$ is an eigenvector of the operator $\mathbf{A}$ with eigenvalue $A_2$ then your operations are correct.

3. Jul 26, 2012

### intervoxel

So

$$\langle A_1|BA|A_2\rangle=\langle A_1|B\rangle \langle A|A_2\rangle= \langle A_1|B\rangle A_2=A_2\langle A_1|B\rangle$$

right?

Thank you.

4. Jul 26, 2012

### The_Duck

No. You are mixing up operators with bras and kets. It looks like A and B are supposed to be operators. If so, the expressions |A>, <A|, |B>, or <B| make no sense. Operators are neither bras nor kets.